(i) Follow Raoult’s law

(ii) They can be separated by fractional distillation

$0.1 M$ glucose solution. Freezing point depression, being a colligative properly, is therefore, nearly twice for solution than for glucose solution of same molarity.

Partial pressure of gas above the solution

$=k_{H} \times$ mole fraction of the gas in the solution

or $\quad p_{A}=k_{H} \times x_{A^{\prime}}$ where $k_{H}=$ Henry’s constant.

$H g I_{2}+2 K I \rightarrow K_{2}\left[H g I_{4}\right] \Longrightarrow 2 K^{+}+\left[H g I_{4}\right]^{2-}$

Therefore, the osmotic pressure increases.

Variation of vapour pressure with Temperature showing boiling point of solution is higher than that of pure solvent.

$=1.353 \mathrm{mol} L^{-1}$ or $1.353 \mathrm{M}$

$m=\frac{w_{B}}{M_{B}} \times \frac{100}{w_{A}}=\frac{13}{98} \times \frac{1000}{87}=1.524 \mathrm{mol} / \mathrm{kg}_{\mathrm{or}} 1.524 \mathrm{m}$

Moles of methanol $\left.=\frac{60}{32}=1.875 \text { (Molar mass }=32\right)$

Moles of water $=\frac{120}{18}=6.667$

Total No. of moles = 1.875 + 6.667 = 8.542

Mole fraction of methanol $=\frac{1.875}{8.542}=0.220$

Mole fraction of water $=\frac{6.667}{8.542}=0.780$

$\Rightarrow \frac{0.87}{31.82}=\frac{27}{M_{B}} \times \frac{18}{100} \Rightarrow M_{B}=\frac{27 \times 18 \times 31.82}{0.87 \times 100}$

$\Rightarrow 177.75 g \mathrm{mol}^{-1}$

$\Delta T_{f}=1.86 \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}=\frac{1.86 \times 3}{60} \times \frac{1000}{100}=0.93$

Freezing point of solution $=273 K-0.93=272.07 K$

Solution of acetone and chloroform shows negative deviations from Raoult’s law because of the formation of the $H$ -bonds between acetone and chloroform molecules.

(i) Association : When solute particles undergo association, number of particles become less and molecular mass determined with the help of colligative property will be more.

(ii) Dissociation : It leads to increase in number of particles, therefore, increase in colligative property, therefore, therefore, decrease in molecular weight because colligative property is inversely proportional to molecular weight.

$\pi v=n R T$ …..(i)

$\pi=$ osmotic pressure; $T=$ Temperature for a solution

If $w$ gram of solute is dissolved in $v$ litres of he solution and $M$ is the molecular mass of the solute, then

$n=\frac{w}{M}$

Substituting this value in equation (v) we get,

$\pi v=\frac{w}{M} R T \Rightarrow M=\frac{w R T}{\pi v}$ …..(ii)

Thus, measuring the osmotic pressure of a solution containing grams of the solute in litres of the solution, at temperature the molecular man, of the solute can be calculated using equation (ii).

Observed molecular mass of benzoic acid,

$M_{B}=\frac{1000 K_{f} \times W_{B}}{\Delta T_{f} \times W_{A}}=\frac{1000 \times 4.9 \times 2}{1.62 \times 25}=242$

Calculated molecular mass of benzoic acid

$=72+5+12+32+1=122$

Van’t Hoff factor

$i=\frac{\text { Calculated molecular mass }}{\text { Observed molecular mass }}=\frac{122}{242}=.504$

Let $\alpha$ be the degree of dissociation

\[ 2 C_{6} H_{5} C O O H \rightleftharpoons\left(C_{6} H_{5} C O O H\right)_{2} \]

Total number of moles after association

$=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

$\therefore \quad i=\frac{1-\frac{\alpha}{2}}{1}=.504$

or $\quad \alpha=(1-.504) \times 2=.496 \times 2=.992$

percentage of association $=99.2 \%$

This method is based upon applying pressure on the solution which is just sufficient to prevent the entry of the solvent into the solution through the semi-permeable membrane. The apparatus used by Berkeley and nartley is shown in Figure.

It consists of a porous tube (open at both ends) containing the semi-permeable membrance of copper ferrocyanide. The porous tube is fitted with a reservoir R on one side and a tube T on the other.

The porous tube is filled with the pure solvent so that the level in the tube T stands at the mark M. The porous tube is fitted into an outer vessel made of gun metal. This vessel has a wide tube at the top which is fitted with a frictionless piston and a pressure gauge as shown in figure. The solution under study is taken in the outer gun metal vessel. As a result of osmosis, the level in the tube tends to fall. The pressure applied on the solution by means of the piston which keeps the level in the tube at is a measure of the osmotic pressure and can be read directly on the pressure gauge.

This method is superior to the other methods because of the following reasons :

(a) In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on the membrane.

(b) The concentration of the solution does not change because the entry of the solvent into the solution is prevented by the external pressure.

(c) The time taken for the measurement of osmotic pressure is much less in this method as compared to the other methods.

Relationships between V.P. and mole fraction for ideal solution

$w_{1}=25 g, \Delta T_{f}=1.62 K$

Substituting these values in equation

$M_{2}=\frac{k_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$M_{2}=\frac{4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} \times 2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{25 \mathrm{g} \times 1.62 \mathrm{K}}=241.98 \mathrm{gmol}^{-1}$

Thus, experimental molar mass of benzoic acid in benzene is $=241.98 \mathrm{g} \mathrm{mol}^{-1}$

Now consider the following equilibrium for the acid:

$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ BHA $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)_{2}$

If $x$ represents the degree of association of the solute then we would have $(1-x)$ mol of benzoic acid left in

unassociated form and correspondingly $\frac{x}{2}$ as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is:

$1-x+\frac{x}{2}=1-\frac{x}{2}$

Thus, total number of moles of particles at equilibrium equals Van’t Hoff factor $i$

But $i=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 g \mathrm{mol}^{-1}}{241.98 \mathrm{gmol}^{-1}}$

$\Rightarrow \frac{x}{2}=1-\frac{122}{241.98}=1-0.504=0.496$

$\Rightarrow x=2 \times 0.496=0.992$

Therefore, degree of association of benzoic acid in benzene is $99.2 \%$

Contribution by the atom B present in the centre of the body = 1

Thus, the ratio of atoms of A : B = 1 : 1

Formula of unit cell = AB

According to available data,

Distance between nearest metal atom $(2 r)=287 \mathrm{pm}$

Edge length for foc crytal $(a)=\sqrt{2} \times 2 r=405.87 \mathrm{pm}$

No. of atoms per unit cell $(z)=4$

Atomic mass of silver $(M)=107.87 \mathrm{gmol}^{-1}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of silver

$=\frac{4 \times 107.87 g m o l^{-1}}{(405.87)^{3} \times 6.022 \times 10^{23} \mathrm{mol}^{-1} \times 10^{-30} \mathrm{m}^{3}}$

$=10.77 \mathrm{g} \mathrm{cm}^{-3}$

Let the edge length of unit cell $=a$

Volume of unit cell $=a^{3}$

Gram formula mass of $N a C l=23+35.5=585 g m o l^{1}$

No. of particles in foc type unit cell $(Z)=4$

Mass of unit cell $=\frac{Z \times \text { Gramformulamassof } N a C l}{\text { Avogadro’s Number }\left(N_{0}\right)}$

$=\frac{4 \times\left(58.5 g \mathrm{mol}^{-1}\right)}{\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)}=3.886 \times 10^{-22} \mathrm{g}$

Density of unit cell $(\rho)=2.165 g \mathrm{cm}^{-3}$

Now, Density of unit cell $=\frac{\left(3.886 \times 10^{-22} g\right)}{\text { Volume of unit cell }\left(a^{3}\right)}$

volume of unit cell $(a)=\frac{\left(3.886 \times 10^{-22} g\right)}{\left(2.165 g \mathrm{cm}^{-3}\right)}$

$=1.795 \times 10^{-22} \mathrm{cm}^{3}=179.5 \times 10^{-24} \mathrm{cm}^{3}$

Edge length $(\mathrm{a})=\left(179.5 \times 10^{-24} \mathrm{cm}^{3}\right)^{1 / 3}=5.64 \times 10^{-8} \mathrm{cm}$

\[ =564 \times 10^{-10} \mathrm{cm}=564 \mathrm{pm} \]

Step II. Calculation of distance between $N a^{+}$ and $C l^{-}$ ions.

Edge length of $N a^{+} C l^{-}$ unit cell $=2\left(r_{N a^{+}}+r_{C l}^{-}\right)$

$2\left(r_{N a^{+}}+r_{C I^{-}}\right)=564 \mathrm{pm}$

or $\left(r_{N a^{+}}+r_{C l^{-}}\right)=\frac{564}{2}=282 \mathrm{pm}$

$\therefore$ Distance in and ions $=282$ pm

Edge length of

$N a C l(a)=2\left(r_{N a^{+}}+r_{C l}\right)=2 . \times 281=562 \mathrm{pm}$

No. of atoms per unit cell $(Z)=4$

$(\because N a C l$ hasfcc structure)

Density of $N a C l(\rho)=2.165 g \mathrm{cm}^{-3}$

Gram molecular mass of $N a C l(M)=23+35.5$

$=58.5 g \mathrm{mol}^{-1}$

$\therefore$ Avogadro’s Number

$\left(N_{0}\right)=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(562)^{3} \times\left(2.165 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.089 \times 10^{23} \mathrm{mol}^{-1}$

no. of $F^{-}$ ions per unit cell $=2 \times 4=8$

(i) Simple cubic lattice : There are points only at the corners of each unit.

(ii) Face-centred cubic lattice : There are points at the corners as well as at the centre of each of the six faces of the cube.

(iii) Body-centred cubic lattice : There are points at the corners as well as in the body centre of each cube.

$B a_{7} K_{3} B i O_{3}, L a_{1.8} S r_{.2} C u O_{4}, Y B a_{2} C u_{3} O_{7}$ etc.

Radius $(\mathrm{r})=\frac{a}{2 \sqrt{2}}$

$a=r \times 2 \sqrt{2}=125 \times 10^{-12} \times 2 \sqrt{2} m$

$=125 \times 2 \times 1.414 \times 10^{-12}=354 \times 10^{-12} \mathrm{m}$

(b) Volume of unit cell $(a)^{3}$

$=\left(354 \times 10^{-12}\right)^{3} m^{3}=4.436 \times 10^{-29} m^{3}$

No. of unit cells in $1.0 \mathrm{m}^{3}$ of $\mathrm{Al}$

$=\frac{\left(1.0 m^{3}\right)}{\left(4.436 \times 10^{-29} m^{3}\right)}=2.25 \times 10^{28}$

According to available data,

Edge length $(a)=564$ pm

Molar mass of $N a C l(M)=23+35.5=58.5 \mathrm{g} \mathrm{mol}^{-1}$

No. of atoms per unit cell” $(Z)=4$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of unit cell

$=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(564)^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=2.16 g \mathrm{cm}^{-3}$

or $\quad a^{3}=\frac{Z \times M}{\rho \times N_{0} \times 10^{-30}}$

According to available data,

No. of atoms in the b.c.c. unit cell $(Z)=2$

Density of unit cell $(\rho)=4.0 \mathrm{g} \mathrm{cm}^{-3}$

Molar mass of $\operatorname{CsCl}(M)=168.5 \mathrm{g} \mathrm{mol}^{-1}$

Avogadro’s no. $\left(N_{0}\right)=6.02 \times 10^{23} \mathrm{mol}^{-1}$

By substituting the values,

$a^{3}=\frac{2 \times\left(168.5 g m o l^{-1}\right)}{\left(4.0 g c m^{-3}\right) \times\left(6.02 \times 10^{23} \mathrm{mol}^{-1}\right) \times 10^{-30}}$

$=1.399 \times 10^{8}(p m)^{3}$

Edge length $(a)=\left[1.399 \times 10^{8}(p m)^{3}\right]^{1 / 3}$

(.: Value of a is in pm)

$=1.12 \times 10^{2} \mathrm{pm}$

(b) $\quad$ (i) In cubic close-packed structures each atom is in direct contact with 12 other atoms. Hence, its coordination number is 12

(ii) In body centred cubic structure coordination number is $8 .$

Methane < Diethyl ether < Ethyl alcohol < Water

Water having highest melting point has the strongest interparticle forces whereas methane having lowest melting point has the weakest intermolecular forces.

The intermolecular forces in these molecules are in the order :

Methane < Diethyl ether < Ethyl alcohol < Water

(b) A crystal lattice may be defined as a regular three dimensional arrangement of identical points in space. While a unit cell may be defined as a three dimensional group of lattice points that generates the whole lattice by translation or stacking.

(c) The vacant space between four touching sphere is called tetrahedral void. Centres of these spheres occupy corners of regular tetrahedron.

The interstitial void formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres, centres of which occupy corners of a regular octahedron.

Conduction of electricity in semi-conductors

The electric conductivity of semi conductors is explained with the help of band theory. As we know, there is only a small energy gap between valence band (filled band) and conduction band (empty band) in semi-conductors. Therefore, some electrons can jump from valence band to vacant band and electric conductance is thus, possible. It can further increases with the rise in temperature since more electrons can jump to the conduction band.

Intrinsic semi-conductors. The extrinsic semi-conductors are formed when impurities of certain elements are added to the insulators.

This process known as doping makes available electrons or holes for conductivity and thus, leads to two types of semi-conduction. These are called n-type semi-conductors and p-type semi-conductors.

n-type semi-conductors. n-type semiconductors are formed when impurity atoms containing more valence electrons than the atoms of the parent in insulator are introduced in it. These are called electron rich impurities. For example, when traces of phosphorus (a group 15 element) is added to pure silicon (a group 14 element) on electron of each phosphorus atom (has five valence electrons) becomes free because it is not involved in any bonding with silicon atom (has four valence electrons). The unbonded electrons are free to carry the electric current. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor.

(ii) p-type semi-conductors. These are formed when impurity atoms containing lesser number of valence electrons than the atoms of the parent insulator element added to it. These are called electron deficient impurities. For example, when traces of boron. (a group 13 element) are added to pure silicon (a group 14 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of boron. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes are responsible for electrical conductivity and will result in p-type semi-conductors.

(ii) Boron, $\mathrm{B}$ is an element of group- 13 and has $2 s^{2} 2 p^{1}$ configuration. It is doped with $S i,$ an element of group-

14 having $3 s^{2} 3 p^{2}$ configuration. All three electrons of boron get bonded with 3 out of 4 electrons of Si and $4^{\text {th }}$ electron of impurity atom (i.e., Si) is responsible for conductivity. Thus it is a n-type semiconductor.

or $M=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{Z}$

According to available data,

Edge length $(a)=288$ pm

No. of atoms per unit cell $(Z)=2$

$(\because \text { bcc structure })$

Density of the element $=7.3 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s No. $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$=52.51 g \mathrm{mol}^{-1}$

Or

$N_{0}=\frac{Z \times M}{a^{3} \times \rho \times 10^{-30}}$

Edge length of the unit cell $(\mathrm{a})=288.4 \mathrm{pm}$

No. of atoms per unit cell $(Z)=2$

Density of the unit cell $(\rho)=7.2 \mathrm{g} / \mathrm{cm}^{3}$

Atomic mass of the element $(\mathrm{M})=52 \mathrm{g} \mathrm{mol}^{-1}$

$N_{0}=\frac{2 \times\left(52 g m o l^{-1}\right)}{(288.4)^{3} \times\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.04 \times 10^{23} \mathrm{mol}^{-1}$

or $\quad Z=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{M}$

Gram atomic mass of $C r(M)=52.0 \mathrm{gmol}^{-1}$

Edge length of unit cell $(\mathrm{a})=289 \mathrm{pm}$

Density of unit cell $(\rho)=7.2 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$\therefore Z=\frac{\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(289^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)\right.}{\left(520 \mathrm{gmol}^{1}\right)}=2$

since the unit cell has 2 atoms, it is body centre in nature.

In a simple cubic unit cell there is only one atom per unit cell.

Let a be the edge length of the unit cell and r be the radius of sphere.

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

As the spheres at the corners are touching each other, the edge length a is equal to $2 \mathrm{r}$.

Volume of the cube $=a^{3}=(2 r)^{3}=8 r^{3}$

$\%$ of the space occupied by spheres

$=\frac{\text { Volume of sphere }}{\text { Volume of cube }} \times 100=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100$

Thus, packing efficiency of simple cubic lattice is 52.4%

(b) Packing Efficiency of Body Centred Cubic (BCC) Structure

We know that in a body centred cubic unit cell there are two spheres per unit cell. Let a be the side of the unit cell and r be the radius of sphere.

volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume occupied by two spheres $=2 \times \frac{4}{3} \pi r^{3}=\frac{8}{3} \pi r^{3}$

Face diagonal $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2 a^{2}}$

In right angled triangle ACD

Body diagonal $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{2 a^{2}+a^{2}}=\sqrt{3} a$

In a body centred cubic unit cell, the spheres at the corners are not touching each other but are in contact with the sphere at the centre. As a result the body diagonal AD is equal to the four times the radius of the sphere.

$\therefore A D=\sqrt{3} a=4 r$

$a=\frac{4}{\sqrt{3}} r$

Volume of unit cell $=a^{3}=\left(\frac{4}{\sqrt{3}} r\right)^{3}=\frac{64}{3 \sqrt{3}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100$

$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100=\frac{8}{3} \times \frac{22}{7} \times \frac{3 \sqrt{3}}{64} \times 100=68 \%$

Thus, packing efficiency of a bcc arrangement is $68 \% .$

(c) Face centred cubic

In a cubic close packing, the unit cell is face centred cube.

In a face centred cubic unit cell there are 4 spheres per unit cell.

Let r be the radius of sphere and a be the edge length of the cube.

Volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume of four spheres $=4 \times \frac{4}{3} \pi r^{3}=\frac{16}{3} \pi r^{3}$

In a face centred cubic unit cell, the spheres at corners are in contact with sphere at the face centre but are not touching each other. Therefore, face diagonal $\mathrm{AC}$ is equal to four times the radius of sphere.

$A C=4 r$

But in right angled triangle ABC (figure)

$A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$

$\therefore \sqrt{2} a=4 r$

$a=\frac{4 r}{\sqrt{2}}$

Volume of the unit cell $=a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}=\frac{64}{2 \sqrt{2}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100=\frac{\frac{16}{3} \pi^{3}}{\frac{64}{2 \sqrt{2}} r^{3}} \times 100$

$=\frac{16}{3} \times \frac{22}{7} \times \frac{2 \sqrt{2}}{64} \times 100=74 \%$

Thus, packing efficiency of ccp arrangement is 74%. Similarly, packing efficiency of hcp is also 74%.

$\mathrm{NaCl}$ has one defect in $10^{15}$ lattice sites at room temperature. The number of defects increases with increases in temperature.

The number of defects increases to one in $10^{6}$ sites at $775 \mathrm{K}$

and one in $10^{4}$ sites at $1075 \mathrm{K} .$ The presence of large number of Schottky defects in crystal results in significant decrease in its density. This defect is also known as vacancy defect.

(ii) Frenkel Defect. This type of defect is created when an ion leaves its correct lattice site and occupies an interstitial site (fig. Frenkel defects are common in ionic compound which have low co-ordination number and in which there is large difference in size between positive and negative ions. For example,

$Z n S, A g C l, A g B r$ and $A g I$

(iii) Interstitial Defects. This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal. The ions occupying the interstitial sites are called interstitials. The formation of interstitial defects in determined by the size of the interstitial ion.

(iv) The “holes’ occupied by electrons are called F-centres (or colour centres) and are responsible for the colour of the compound and many other interesting properties. For example, the excess sodium in $\mathrm{NaCl}$ makes the crystal appear yellow, excess potassium in $K C l$ makes it violet and excess lithium in LiClmakes it pink. Greater the number of F-centres, greater is the intensity of colour. Solids containing F-centres are paramagnetic because the electrons occupying the “holes’ are unpaired.

Face diagonal $=4 \times r=4 \times 125 \mathrm{pm}=500 \mathrm{pm}$

But face diagonal $=\sqrt{2} \times$ edge length

$\therefore$ Edge length $=\frac{\text { Face diagonal }}{\sqrt{2}}=\frac{500}{\sqrt{2}}=354 \mathrm{pm}$

(b) Volume of one unit cell

$=\mathrm{a}^{3}$

$=\left(3.54 \times 10^{-8} \mathrm{cm}\right)^{3}$

No. of unit cells $1.00 \mathrm{cm}^{3}$

$=\frac{1.00}{\left(3.54 \times 10^{-8}\right)^{3}}$

$=2.26 \times 10^{22}$

mercaptan $\left(C_{2} H_{5} S H\right)$

alkenes with the help of $H_{2}$

conformation is $12.5 \mathrm{kJ} \mathrm{mol}^{-1}$

The order of stability of conformers is in the following order:

Staggered $>$ gauche (skew) $>$ eclipsed.

For Example:

Types of cracking : The process of cracking is of two types:

(i) Thermal cracking.

(ii) Catalytic cracking.

(i) Thermal cracking: When cracking is carried at high temperature, it is called thermal cracking.

(ii) Catalytic Cracking: When cracking is carried at low temperature in the presence of a catalyst such as alumina or silica, is called catalytic cracking.

Each $1^{\circ}$ Carbon is attached to 3 hydrogen, Each $2^{\circ}$ carbon is attached to 2 hydrogen atoms. Each tertiary ‘C’ atom is bonded to one and quaternary $^{\epsilon} \mathrm{C}^{\prime}$ is not attached to any hydrogen atom.

All five Carbon atoms are attached to total 15 hydrogen atoms.

TwoCarbon atoms are attached to total 4 hydrogen atoms.

OneCarbon atom is attached to total one hydrogen atom

OneCarbon atom is not attached to any hydrogen atom.

(i), (ii) and (iii) Newman eclipsed projection of butane.

(ii) 3 methyl pent $-1$ -ene- 4 -yne.

It should follow Huckel’s Rule, i.e., in a conjugated, planar, cyclic system if number of delocalized electrons is $(4 n+2) \pi,^{c} n^{\prime}$ being integer, $i . e ., 1,2,3,$ etc. it will be aromatic, otherwise not, e.g.

Benzene, naphthalene, anthracene have $6,10,14$ electrons in conjugated, planar, cyclic system.

These products prove the structure I and II (Kekule structures of benzene).

In Benzene, each carbon is $s p^{2}$ hybridised carbon $(33 \%$ of character), which is less electronegative than hybridised carbon, therefore, it is less acidic.

In hexane, each $^{c} C^{\prime}$ is hybridised carbon $(25 \%$ s-character), which is least electronegative, therefore, it is least acidic.

since electron density is maximum at $o$ and $p$ -positions $S O_{3}$ will attack at $o$ and $p$ -positions.

$\mathrm{NO}_{2}$ will attack at $o$ and $p$ -positions because electron density is maximum at ortho and $p$ -positions.

(ii) When acetylene is passed through ammonical silver nitrate

solution, white precipitate of silver acetylide will be formed.

(iii) When chlorobenzene is condensed with chloral, in presence of conc. $H_{2} S O_{4},$ DDT is formed.

At ordinary pressure butane $\left(C_{4} H_{10}\right),$ Propane $\left(C_{3} H_{8}\right)$ and ethane $\left(C_{2} H_{6}\right)$ are gases. Under high pressure, they are easily liquefied. The petroleum gas which has been liquefied under pressure is called liquefied petroleum gas (LPG). The domestic gas cylinder contains LPG.

LPG a better domestic fuel : Components of LPG burn easily and produce a lot of heat, so LPG is used as a good domestic fuel. Due to the following qualities, LPG is considered as a better fuel.

(i) $\mathrm{LPG}$ is an efficient fuel as it has high calorific value, burning of 1 gram of LPG in air produces about 50 kilo joule of heat energy.

(ii) LPG burns with a smokeless flame, so causes no pollution.

(iii) It leaves no ash on burning so it is a clean fuel.

(iv) On burning no poisonous gases are formed.

(v) It is easy to handle and convenient in storing.

Principle of refining of petroleum :”The process of separating different components of petroleum, is known as refining. It is carried out by the process of fractional distillation which is based on the fact that different components of petroleum have different boiling point ranges.

The crude-oil is heated in a tank to a temperature of about and all the vapours formed are fed into a tall fractionating column near the bottom. Except asphalt, all the other components are in vapour state. As the mixture of hot vapours rises up in the column, it begins to cool The vapours of components having higher boiling point-range condense first in the lower part and vapours of components with lower boiling point-range condense later at the upper part of column. Therefore, the components with the highest boiling point condenses first and is collected in the lowest part of the column and the component having lowest boiling point condenses last and is collected in the top most part of column. The vapours that do not liquify, are taken out from the top of the column. This component is called petroleum-gas. The residual oil or liquid residue which does not vapourise in the column is called residual oil. It is collected in the base of column and subjected to further distillation.

Products of petroleum : The various components obtained at different heights of fractional column, (in order of top to bottom of column) are as follows :

-acid reacts with alkenes to gives addition products

(ii) Reaction with halogen : Halogens (chlorine and bromine only react at ordinary temperature and without exposure to $U V$ light to form vicinal halide.

(iii) Oxidation

(a) Oxidation with dilute aqueous potassium permanganate solution : At low temperature alkenes form vicinal glycols.

[This reaction is called hydroxylation as hydroxyl groups are attached across the double bond. $]$

(b) With alkaline potassium permanganate at higher temperature : In such conditions cleavage of $C=C$ bond occurs forming carboxylic acids.

(iv) Ozone: With ozone alkenes form ozonides which undergo cleavage to produce carbonyl compounds. This method was extensively used to determine the position of double bond in alkenes in past.

Procedure : Ozone is passed through a solution of an alkene in inert solvent such as $\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}$ or $\mathrm{CCl}_{4}$ at temperature $196-200 K .$ Alkenes are oxidised to ozonised which are not usually isolated but reduced in situ, with zinc dust and water or $H_{2} / P d$ to give carbonyl compound. The location of double bond is assigned on the basis of products.

(a) In case of symmetrical alkene, only one aldehyde is formed on reduction with zinc and water:

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}+\mathrm{O}_{3} \longrightarrow$

(b) If double bond is terminal, methanal is one product and a higher aldehyde is other product.

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{O}_{3} \longrightarrow$

(c) If ketone is formed, branching is there at C atom having double bond.

(ii) Functional isomerism.

The acid produced in the reaction acts as catalyst (autocatalyst) for the reaction. Hence, the reaction becomes faster after some time.

Demulsification can be done by boiling or freezing.

According to Le Chatelier’s principle, if we increase the temperature, equilibrium will shift in the backward direction, i.e., gas is released from the adsorbed surface.

When oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, coagulation of both the sols occurs.

A distinction can be made between absorption and adsorption by taking an example of water vapour. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

(i) Double decomposition : A colloidal sol of arsenious sulphide is obtained by passing hydorgen sulphide into a solution of arsenious oxide in distilled water.

$A s_{2} O_{3}+3 H_{2} S \rightarrow A s_{2} S_{3}+3 H_{2} O$

(ii) Oxidation : A colloidal solution of sulphur can be obtained by passing hydrogen sulphide into a solution of sulphur dioxide in water through a solution of an oxidizing agent (bromide water, nitric acid etc.)

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \rightarrow H_{2} O+S$

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \longrightarrow 3 \mathrm{S}+\mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \longrightarrow H_{2} O+[S]$

Preparation of platinum sol : It is prepared by electrical disintegration or Bredig’s arc method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vaporizes some of the metal, which then condenses to form particles of calloidal size.

(i) Manufacture of $N H_{3}$ from $N_{2}$ and $H_{2}$ by Haber’s process using iron as catalyst.

$N_{2}+3 H_{2} \stackrel{F e}{\longrightarrow} 2 N H_{3}$

(ii) Manufacture of $\mathrm{CH}_{3} \mathrm{OH}$ from $\mathrm{CO}$ and using $(\mathrm{a}$ mixture of $C u, Z n O$ and $C r_{2} O_{3}$

$\mathrm{CO}+2 \mathrm{H}_{2} \frac{\mathrm{Cu}+\mathrm{ZnO}+\mathrm{Cr}_{2} \mathrm{O}_{3}}{\mathrm{Heat}} \rightarrow \mathrm{CH}_{3} \mathrm{OH}$

(iii) Oxidation of with using as catalyst in ostwald process.

(iv) Hydrogenation of oils to form vegetable ghee using finely divided nickel.

$\mathrm{Oil}+\mathrm{H}_{2} \stackrel{\mathrm{Ni}}{\longrightarrow}$ Vegetable ghee.

Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. An important zeolite catalyst used in the petroleum industry is ZSM- – . It converts alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

(1) In medicines : A wide variety of pharmaceutical preparations are emulsions. For example, emulsions of cod liver oil. These emulsified oils are easily acted upon by digestive juices in the stomach and, hence are readily digested.

(2) Digestion of fats : Digestion of fats in the intestines is aided by emulsification.

(3) In disinfectants : The disinfectants such as dettol give emulsions of the oil in water type when mixed with water.

(4) In building roads : An emulsion of asphalt and water is used for building roads. In this way there is no necessity of melting the asphalt.

(i) Nature of gas: Under given conditions of temperature and pressure, the easily liquifiable gases like $N H_{3}, H C l, C O_{2},$ etc. are adsorbed to a greater extent than

the permanent gases such as $H_{2}, O_{2}, N_{2}$ etc. It is because the vander waal’s forces or molecular forces are more predominant in the former than in later category.

Effect of Nature of the Adsorbent: Activated charcoal (ii) is the most common adsorbent for the gases which can be easily liquefied. The gases $H_{2}, O_{2}$ and $N_{2}$ are adsorbed on metals like $N i, P d$ etc.

(iii) Specific area of the solid : Specific area of an adsorbing solid is the surface available for adsorption per gram of the adsorbent. Greater the specific area of the solid, greater would be its adsorbent power.

(vi) Effect of pressure : It is expected that extent of adsorption increases with increase of pressure.

(v) Effect of temperature : In physisorption

$x / m$ decreases with rise in temperature. While in chemisorptionslightly increases in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that, like chemical reactions, chemisorption also requires activation energy.

(i) The mathematical expressions representing adsorption are:

Freundlich adsorption isotherm: $\frac{x}{m}=k p^{1 / n}$

Langmuir adsorption isotherm: $\frac{x}{m}=\frac{a p}{1+b p}$

(ii) Langmuir adsorption isotherm is more general and is applicable at all pressure while Freundlich adsorption isotherm fails at high pressure. The latter can be deduced from the former.

(iii) Langmuir adsorption postulates that adsorption is monomolecular whereas Freundlich adsorption suggests that it is multimolecular.

$\bullet$ By finely dividing the adsorbent.

$\bullet$ By removing the gases already adsorbed

$\bullet$ By making the surface of adsorbent rough by chemical or mechanical methods.

extent of adsorption $\left(\frac{x}{m}\right)$ and the pressure $p$ of the gas at constant temperature is called adsorption isotherm.

. Effect of temperature : Magnitude of adsorption decrease with increase in the temperature.

A graph drawn between extent of adsorption $\left(\frac{x}{m}\right)$ and temperature ( $T$ ) at constant pressure is called adsorption isobar physical adsorption isobar shows a decrease in If with rise in temperature, the isobar of chemisorption $x / 1$ shows an increase in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that like chemical reactions, chemisorption also requires activation energy.

(ii) Lyophobic colloids : The word ‘lyophobic’ means liquid-hating. Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloids. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence, are not stable. Further, once precipitated, they do not give back the colloidal sol by simple addition of the dispersion medium. Hence, these sols are also called irreversible sols. Lyophobic sols need stabilising agents for their preservation.

Difference between associated colloids multimolecular and macromolecules colloids. Multimolecular colloids are formed by the aggregation of a large number of small atoms molecules such as S8. Macromolecular colloids contain molecules of large size like starch the size of molecules in this case have the dimensions of colloids. Their size lies in the colloidal range. The associated colloids are formed by electrolytes which dissociate into ions and these ions associate together to form ionic micelles whose size lies in the colloidal range, soap is common example of this type . The micelle formation occurs above a particular concentration (called critical micellisation concentration) and above a particular temperature, called Kraft temperature.

(ii) The constituent liquids of the emulsion separate out. In other words, demulsification occurs.

(iii) On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

(iv) Scattering of light by the colloidal particles takes place and the path of light becomes illuminated. This is called Tyndall effect.

(2) Electrophoresis : In this process colloidal particles move towards oppositely charged electrodes, get discharged and precipitated.

(3) Coagulation : The stability of the lyophobic sol is due to the presence of charge on colloidal particles. If, some how, the charge is removed, the particles will come nearer to each other to form aggregate (or coagulate) and settle down.

(2) Aerosol : It is a colloidal dispersion of a liquid in a gas, common example is, fog.

(3) Hydrosol :It is a colloidal sol of a solid in water as the dispersion medium, common example is starch sol or gold sol.

(i) Oil dispersed in water (O/W type) and

(ii) Water dispersed in oil (W/O type ).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

Emulsions of oil in water are unstable and sometimes they separate into two layers on standing. For stabilisation of an emulsion, a third component called emulsifying agent is usually added. The emulsifying agent forms an interfacial film between suspended particles and the medium. The principal emulsifying agents of O/W emulsions are proteins, gums, natural and synthetic soaps, etc., and for W/O, heavy metal, salts of fatty acids, long chain alcohols, etc.

Emulsions can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed, forms a separate layer. The droplets in emulsions are often negatively charged and can be precipitated by electrolytes. They also show Brownian movement and Tyndall effect. Emulsions can be broken into constituent liquids by heating, freezing, etc.

Explanation : The cleansing action of soap can be explained keeping in mind that a soap molecule contains a non-polar hydrophobic group and a polar hydrophilic group. The dirt is held on the surface of clothes by the oil or grease which is present there. since oil or grease are not soluble in water, therefore, the dirt particles cannot be removed by simply washing the cloth with water. When soap is applied, the non-polar alkyl group dissolves in oil droplets while the polar $-C O O^{-} N a^{+}$ groups remain dissolved in water (Figure). In this way, each oil droplet is surrounded by negative charge. These negatively charged oil droplets cannot coagulate and a stable emulsion is formed. These oil droplets (containing dirt particles) can be washed away with water along with dirt particles.